Expected values in infinite square well

[Aikon]

Ok...this must sound stupid, because i didn't found answer on the web and on my books...but i am having trouble with the infinite square well.
I want to calculate <x>.
V(x)=0 for 0<=x<=a
<br /> &lt;x&gt;=\frac{2}{a}\int^{a}_{0} x \sin^2(\frac{n\pi}{a}x)dx<br />
Doing integration by parts i got to:
\frac{2}{a}\left[\frac{a^2}{2n}-\int^a_0\frac{a}{2n}dx\right]=0
What i am doing wrong?
Thank you

 

[mfb]

I would expect an error in your integration, as the first expression is not zero and the following one is.

 

[tadchem]

<br /> &lt;x&gt;=\frac{2}{a}\int^{a}_{0} x \sin^2(\frac{n\pi}{a}x)dx<br />

set
<br /> u=\frac{n\pi}{a}x<br />

look up
<br /> \int u \sin^2(u)du <br />
in a Table of Integrals to find
<br /> \frac{u^2}{4}-\frac{u\sin(2u)}{4}-\frac{cos(2u)}{8}<br />

 

[Aikon]

set
<br /> u=\frac{n\pi}{a}x<br />

look up
<br /> \int u \sin^2(u)du <br />
in a Table of Integrals to find
<br /> \frac{u^2}{4}-\frac{u\sin(2u)}{4}-\frac{cos(2u)}{8}<br />

Yeah, i got it...it is easier to use table of integrals.
it gives the expected &lt;x&gt;=\frac{a}{2} for any value of n.
thanks